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If $f(x)$ is defined on $(0,1)$,then prove that the domain of definition of $f(e^x)+f(\ln|x|)$ is $(-e,-1)$


As given in the question,$0<x<1$
$\Rightarrow 1<e^x<e$
$f(1)<f(e^x)<f(e)$
Similarly,$0<x<1$
$-\infty<\ln|x|<0$
$f(-\infty)<f(\ln|x|)<f(0)$
But i do not know how to solve it further and reach to the desired answer.Is my approach correct?If not,what is the correct method.Please help me.Thanks.

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  • $\begingroup$ Figure out the $x$ such that $e^x \in (0,1)$. Figure out the $x$ such that $\ln |x| \in (0,1)$. Take the intersection. $\endgroup$
    – copper.hat
    Dec 9, 2015 at 17:13

2 Answers 2

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You need to find for which $x$ are $e^x$ and $\ln(|x|)$ both in $(0,1)$. Since $e^x$ must be in $(0,1)$, it must be that $x<0$. So the question is for which $x<0$ is $\ln(|x|)\in(0,1)$. Since $x<0$ this is the same as $\ln(-x)\in(0,1)$. Look at the graph of $\ln(-x)$ for $x<0$, it's strictly decreasing and is $1$ when $x=-e$ and is $0$ when $x=-1$, thus $\ln(|x|)$ is in $(0,1)$ exactly when $x$ is between $-e$ and $-1$.

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Here is the issue. $f(x)$ is defined on $(0,1)$. What may be confusing you is the fact that $x$ is the same symbol used to make this statement, and again in the definition of a new function $g(x) = f(e^x) + f(\ln|x|)$.

Instead, think of it this way: $f(y)$ is defined for $y \in (0,1)$. Now, we're defining $g(x) = f[h_1(x)] + f[h_2(x)]$, where $h_1(x) \equiv e^x$ and $h_2(x) \equiv \ln|x|$. So, the definition of $f$ requires that $h_1(x) \in (0,1)$ and that $h_2(x) \in (0,1)$, rather than requiring that necessarily $x \in (0,1)$. With this in mind, you then need to determine what this implies for $x$ itself in the definition of $g(x)$, which should be straightforward.

I hope this helps.

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